Results 1 to 6 of 6

Math Help - Challenging Limit Question!

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    45

    Challenging Limit Question!

    Ok I've got to evaluate the following limit:

    But I've got no idea how the absolute values can be rewritten in this limit?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Kataangel View Post
    Ok I've got to evaluate the following limit:

    But I've got no idea how the absolute values can be rewritten in this limit?
    Note that \left|x-1\right|=\left\{\begin{array}{rl} x-1, & \text{ if }x\geq 1\\ {\color{red}-x+1}, & \text{ if }x<1\end{array}\right. and \left|x+1\right|=\left\{\begin{array}{rl} {\color{red}x+1}, & \text{ if }x\geq -1\\ -x-1, & \text{ if }x<-1\end{array}\right.

    Thus, \lim_{x\to 0}\frac{x}{\left|x-1\right|-\left|x+1\right|}=\lim_{x\to 0}\frac{x}{(-x+1)-(x+1)}=\lim_{x\to0}-\frac{x}{2x}.

    I'm sure you can take it from here.
    Last edited by Chris L T521; October 4th 2009 at 10:21 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2008
    Posts
    45
    Quote Originally Posted by Chris L T521 View Post
    Note that \left|x-1\right|=\left\{\begin{array}{rl} x-1, & \text{ if }x\geq 1\\ {\color{red}-x-1}, & \text{ if }x<1\end{array}\right. and \left|x+1\right|=\left\{\begin{array}{rl} {\color{red}x+1}, & \text{ if }x\geq -1\\ -x+1, & \text{ if }x<-1\end{array}\right.

    Thus, \lim_{x\to 0}\frac{x}{\left|x-1\right|-\left|x+1\right|}=\lim_{x\to 0}\frac{x}{(-x-1)-(x-1)}=\lim_{x\to0}-\frac{x}{2x}.

    I'm sure you can take it from here.
    Thanks very much for illustrating Chris, but I have no idea how you chose x-1 in the 2nd set of brackets in \lim_{x\to 0}\frac{x}{(-x-1)-(x-1)}???

    And why the red ones in particular? I'm not sure which set of arbitrary values this limit should be taking? This would be easier with one sided limits.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Kataangel View Post
    Thanks very much for illustrating Chris, but I have no idea how you chose x-1 in the 2nd set of brackets in \lim_{x\to 0}\frac{x}{(-x-1)-(x-1)}???

    And why the red ones in particular? I'm not sure which set of arbitrary values this limit should be taking? This would be easier with one sided limits.
    erm...woops. I took the wrong one from each group (and had a couple typos). Please note my correction.

    I looked at the intervals of each piecewise term, and picked the ones that contained zero in the interval (since we're taking the limit as x approaches zero).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2008
    Posts
    45
    By the way, how do you know which of the greater than or less than signs you should put the equals sign under?

    i.e. how can you tell which sign needs to have the 'or equal' with it because attaching the equals under greater than or less than are equally possible?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Kataangel View Post
    By the way, how do you know which of the greater than or less than signs you should put the equals sign under?

    i.e. how can you tell which sign needs to have the 'or equal' with it because attaching the equals under greater than or less than are equally possible?
    It really doesn't matter which one. Its just a matter of preference, I guess.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A Challenging PDE question
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: August 30th 2011, 05:18 AM
  2. A Challenging Question
    Posted in the Advanced Statistics Forum
    Replies: 10
    Last Post: August 15th 2009, 06:03 PM
  3. Challenging question?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 1st 2008, 02:34 AM
  4. Challenging--- Central Limit Theorem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 22nd 2008, 07:12 PM
  5. Please help... challenging Integral Question
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 20th 2007, 06:06 PM

Search Tags


/mathhelpforum @mathhelpforum